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Description: Simion and Schmidt showed in 1985 that the cardinality of the set Sn(123,132) length n permutations avoiding the patterns 123 and 132, is 2n-1, but in the other side 2n-1 is the cardinality of the ...

Simion and Schmidt showed in 1985 that the cardinality of the set Sn(123,132) length n permutations avoiding the patterns 123 and 132, is 2n-1, but in the other side 2n-1 is the cardinality of the set Bn-1 = {0,1}n-1 of length (n-1) binary strings. Theoretically, it must exist a bijection between Sn(123,132) and Bn-1. In this paper we give a constructive bijection between Bn-1 and Sn(123,132); we show that it is actually an isomorphism and illustrate this by constructing a Gray code for Sn(123,132) from a known similar result for Bn-1. As we noted that an isomorphism between two combinatorial classes is a closeness preserving bijection between those classes, that is, two objects in a class are closed if and only if their images by this bijection are also closed. Often, as in this paper, closeness is expressed in terms of Hamming distance. Isomorphism allows us to find out some properties of a combinatorial class X (or for the graph induced by the class X) if those properties are found in the pre image of the combinatorial class X; some mentioned properties are hamiltonian path, graph diameter, exhaustive and random generation, and ranking and unranking algorithms.

Bijection and Isomorphism on Graph of Sn(123,132) from One of (n-1) Length Binary Strings
A. Juarna1, A.B. Mutiara2
Faculty of Computer Science and Information Technology, Gunadarma University Jl. Margonda Raya No.100, Depok 16424, Indonesia 1,2 {ajuarna,amutiara}@staff.gunadarma.ac.id piecewise comparison to 231), while permutation 4321 ∈ S4(T) since it not contain any subsequence which is piecewise comparison to any pattern of T. Also s3(123) = 5 because S3(123) = {132, 213, 231, 312, 321}. Fundamental questions about pattern-avoiding permutations problems are: to determine sn(T) viewed as a function of n for given T, to find an explicit bijection (a one-to-one and onto correspondence) between Sn(T) and Sn(T’) if sn(T) = sn(T’), and 3. to find relations between Sn(T) and other combinatorial structures. By determining sn(T) we mean finding explicit formula, or ordinary or exponential generating functions. From these researches, a number of enumerative results have been proved, new bijections found, and connections to other fields established. 1. 2. Problems of pattern avoiding permutations appeared for the first time when Knuth [5], in his text book, posed a sorting problem using single stack. This problem actually is the 312patterns avoiding permutations. In the other section of his book, he showed that the cardinality of all three-lengthpatterns-avoiding permutations is the Catalan numbers. Investigations on problems of pattern avoiding permutations then become wider to some set of patterns of length three, four, five, and so on, some combinations of these patterns, generalized patterns, and permutations avoiding some patterns while in the same time containing exactly a numbers of other patterns. Pattern avoiding permutations have been proved as useful language in a variety of seemingly unrelated problems, from theory of Kazhdan-Lusztig polynomials, to singularities of Schubert varieties, to Chebyshev polynomials, to rook polynomials for a rectangular board, to various sorting algorithms, sorting stacks and sortable permutations [4], statistic permutation [6], also in practical application such as on cryptanalysis (see [7] for example). The first systematic study of patterns avoiding permutations undertaken in 1985 when Simion and Schmidt [9] solved the problem with patterns come from every subset of S3. The idea of this paper is the following propositions, Proposition 1 (see [9]) The number of (123,132)-avoiding permutations in Sn, n ≥ 1 is sn(123,132) = 2n-1.

Abstract—Simion and Schmidt showed in 1985 that the cardinality of the set Sn(123,132) length n permutations avoiding the patterns 123 and 132, is 2n-1, but in the other side 2n-1 is the cardinality of the set Bn-1 = {0,1}n-1 of length (n-1) binary strings. Theoretically, it must exist a bijection between Sn(123,132) and Bn-1. In this paper we give a constructive bijection between Bn-1 and Sn(123,132); we show that it is actually an isomorphism and illustrate this by constructing a Gray code for Sn(123,132) from a known similar result for Bn-1. As we noted that an isomorphism between two combinatorial classes is a closeness preserving bijection between those classes, that is, two objects in a class are closed if and only if their images by this bijection are also closed. Often, as in this paper, closeness is expressed in terms of Hamming distance. Isomorphism allows us to find out some properties of a combinatorial class X (or for the graph induced by the class X) if those properties are found in the pre image of the combinatorial class X; some mentioned properties are hamiltonian path, graph diameter, exhaustive and random generation, and ranking and unranking algorithms. Keywords-pattern-avoiding permutations; binary strings, constructive bijection; Hamming distance; combinatorial isomorphism.

I.

INTRODUCTION

In this paper an element denotes a member of a list or set, and a term denotes a term in a string or sequence. Let x = x1 x2 ... xn and y = y1 y2 ... yn be two strings of same length. We say x and y are piecewise comparison if xi ≤ xj whenever yi ≤ yj. Let [n] be the set of all non-negative integers less than or equal to n. We denote by Sn the set of all permutations of [n] and its cardinality is obviously n!. Let π ∈ Sn and τ ∈ Sk be two permutations, k ≤ n. We say π contains τ if there exists k integers 1 ≤ i1 < i2 ... ik ≤ n such that subsequence π i Kπ i is piecewise comparison to τ; in such context τ is usually called a pattern. We say that π avoids τ, or π is τ-avoiding, if such subsequence does not exist. The set of all τ-avoiding permutations in Sn is denoted by Sn(τ) and sn(τ) is its cardinality. For an arbitrary finite collection of patterns T, we say π avoids T if π avoids any τ ∈ Sk; the corresponding subset of Sn is denoted by Sn(T) while sn(T) is its cardinality. For examples, let T = {123,231,1324} is a set of patterns. Clearly permutation 1234567 ∉ S7(T) since it contains 123, permutation 652341 ∉ S6(T) since it contain 234 which is piecewise comparison to 123 (and also 231 and 341 which are
1 k

For example, Figure 1 is the matrix representation of permutation 6573421 ∈ S7(123,132). If we trace the terms of π in (1) from the left to the right, at first we will find π1 as the second largest term in π (after n). If we remove π1, then π2 again will be the second largest, and so untilπk-1. Next, πk = n is the largest term of π. This tracing and interpretation is similar for the third part of π until one place before the largest term. Now, we associate π ∈ Sn(123,132) to s, a binary string of length (n-1), and assign the largest of π whenever we find 1 in s and assign the second largest of π whenever we find 0 in s. It is easy to see that this construction is a bijection, so we get the following proposition: Proposition 2 For each n ≥ 1, there exists a constructive bijection between Bn-1 and Sn(123,132). Proof. Let s = s1s2... sn ∈ Bn-1. We construct its corresponding π ∈ Sn(123,132) by determining πi, 1 ≤ i < n, as follows: if Xi = {1, 2, ..., n} – {π1, π2, ..., πi-1}, then set:

The cardinality of set Sn(123,132), as stated by SimionSchmidt, is the number of elements of Bn-1, the set of all binary strings having length (n-1) without any restriction. This paper gives (in the next section) constructive bijection between Bn-1 and Sn(123,132). Then, in section 3 we show that this bijection is actually isomorphism. Remark that is not always the case: a bijection between combinatorial classes may magnify the distance between two consecutive objects. This result allows us to construct in section 4 a Gray code for Sn(123,132). In the final part some concluding remarks are given. II. CONSTRUCTIVE BIJECTION BETWEEN Bn −1 S n (123,132)
AND

Simion and Schmidt proved that cardinality of set Sn(123,132) is 2n-1, but the 2n-1 is also cardinality of Bn-1, set of all binary strings of length n-1. Theoretically it must be exists a bijection between Sn(123,132) and Bn-1; here we construct such a bijection. The general pattern of π ∈ Sn(123,132), as is mentioned in Proposition 1, can be described as three parts as,

πi = ⎨

⎧ largest element in X i if si = 1 ⎩ second largest element in X i if si = 0

Figure 1. π = 6573421 ∈ S7(123,132) consist of three part as is mentioned by (1). Notice that the third part is an element of S4(123,132), the first stage in the verification of π = 6573421 as element of S7(123,132) recursively using (1).

A graph associated with a combinatorial class is a graph where objects of the class act as vertices of the related graph. Two vertices of this graph are connected (or adjacent) if the associated two combinatorial objects are closed, that is fulfill a predetermined condition(s), usually in the term of Hamming

distances. Two graphs G and H are said to be isomorphic if there is a bijection ϕ such that (u,v) is an edge in G if and only if (ϕ(u), ϕ(v)) is an edge in H. Before exploring the graph associated with the combinatorial classes Bn-1 and Sn(123,132) and showing the isomorphism between the two graph, we define the closeness properties of two elements of Bn-1 and Sn(123,132) and then give a theorem concerning the isomorphism. Definition 1 1. Two binary strings Bn-1 are closed if they differ in a single position. 2. Two permutations in Sn(123,132) are closed if they differ by a transposition of two terms. Theorem 1 The bijection (2) is a combinatorial isomorphism, that is, two binary strings in Bn-1 are closed if and only if their images in Sn(123,132) under this bijection are closed. Proof. Let x and x’ be two elements of Bn-1 which differ at position i, and also, without loss of generality, let xi = 1, and: x = x1...xi-110...01xj+1...xn-1 x = x1...xi-100...01xj+1...xn-1 With the contiguous sequence of 0s: xi+1 = xi+1 = ... = xj-1 = 0 eventually empty. • If xj until xn-1 is 0 then πn = (m-1) for π and m for π’. • Let m be the largest element in Xi as is mentioned in (2). Let π, π’ ∈ Sn(123,132) the images of x and x’ by the bijection (2), clearly πi = m, πi+1 = (m-2), and so on, while π1’ = (m-1), π1+1’ = (m-2), and so on. Then the shapes of π and π are:

IV.

GRAY CODE FOR S n (123,132)
DISTANCES

AND THE HAMMING

A binary string is a string over a binary alphabet, {0,1}. The set of binary strings of length p codes the set of nonnegative integers over closed interval [0, 2p-1]. For example, set of all 3 length binary strings is {000, 001, 010, 011, 100, 101, 110, 111} and represents set of all non-negative integers less than or equal to 7, the all non-negative integers over the closed interval [0, 23-1]. A Gray code for binary strings is a listing of all p length p binary strings so that successive strings (including the first and last) differ in exactly one bit position [8]. The simple and bestknown example of Gray code for binary strings is binary reflected Gray code which can be described the following recursive definition:

ε ⎧ Bp = ⎨ 0 ⋅ B p −1 o 1 ⋅ B p −1 ⎩

p=0 p ≥1

(3)

where ε is empty string, α ⋅ B is the list obtained by concatenation α to each string of B , o is concatenation operator of two lists, and B is the list obtained by reversing B. Fist(Bp) = 0p since it is constructed by recursively concatenation 0 to ε and so on in p times, while Last(Bp) = 10p-1 since it just concatenation 1 to First(Bp-1) and since Last( B p ) = First(Bp). For examples, B1 = {0, 1}, B2= {00, 01, 11, 10}, and B3 = {000, 001, 011, 010, 110, 111, 101, 100}. Since the first and last elements of Bp also differ in one bit position, the code is in fact a cycle. Generating of (3) can be implemented efficiently as a loop free algorithm [1]. Note that, since a binary Gray code is a cycle, it can be viewed as a Hamilton cycle in the n-cube. Existence of at least a Hamiltonian cycle in the graph of Sn(123,132), as is showed in the last part of the previous section, is an indication that there is at least a Gray code for Sn(123,132). Since there is a bijection between Bn-1 and Sn(123,132), here we construct a Gray code for Sn(123,132). By considering bijection (2), Gray code Bp (3) is transformed into following Gray code for Sn(123,132):
{1} n =1 ⎧ ⎪ * S n (123,132) = ⎨( n − 1) ⋅ S n −1 (123,132) o ⎪ n ⋅ S n −1 (123,132) n≥2 ⎩
*

π = π1... πi-1 m (m-2) ... (m-j+i+1) (m-1) πj+1... πn-1 πn π’ = π1... πi-1 (m-1) (m-2) ... (m-j+i+1) m πj+1... πn-1 πn
The case for xi = 0 is similar. □ Since (3) is cyclic, we can draw an (n-1)-cube graph of Bn-1 and also we can find at least a Hamiltonian cycle in the graph. And since (2) is an isomorphism, we also can draw a congruent graph of Sn(123,132) and also can find the Hamiltonian cycle. Figure 2 shows the two graphs for n = 4 together with one of their Hamiltonian path.

(4)

Figure 2. Isomorphism between graph B3 and graph S 4 (123,132) . This figure also shows a Hamiltonian cycle in each graph, as is indicated by the arrows. Notice that the Hamiltonian path in S 4 (123,132) is the isomorphic

image of the path in B3

where S n −1 (123,132) is Sn-1(123,132) after replacing (n-1) with n. This replacement is taken place since 0, which is the prefix to the first part of (3), is associated to (n-1), the second largest element as is mentioned in (2). Hence (n-1) must be prefix to the second part of (4). For examples, S2(123,132) = {12, 21}, S3(123,132) = 2⋅{13, 31} o 3⋅{12, 21} = {213, 231, 321, 312}. Table 1. shows the list of B4 together with its image, the list of S5(123,132).

The recursively properties of (4) imply First(Sn(123,132)) = (n-1)(n-2)...1n. In the other hand, since Last ( S n − 1 (123,132)) = First(Sn-1(123,132)), so Last ( S n (123,132)) must be n⋅(n-1)⋅(n3)...1(n-1). Proposition 3. The Hamming distance between two consecutive elements of Sn(123,132) is 2 and, except between the first and the last, the two different terms are adjacent. Proof. For n = 2 the Hamming distance is between 12 and 21 which is 2. For n > 2, Hamming distance between two consecutive elements of Sn(123,132), except between the first and last elements, is determined recursively by the distance in the smaller list, and so on, and finally by the distance in S2(123,132) which is 2. Concatenating (n-1) and n, respectively to the two parts of (4), of course will not change the Hamming distance values in each part. Also, replacing (n* 1) with n in S n −1 (123,132) will not change the Hamming distance between each its two consecutive elements. So we only must to check the Hamming distance between * Last ((n − 1) ⋅ S n −1 (123,132)) and First (n ⋅ S n −1 (123,132)) , as follow: * Last ((n − 1) ⋅ S n −1 (123,132))
* = (n − 1) ⋅ Last ( S n −1 (123,132))

AUTHORS PROFILE A. Juarna is a combinatorlist at Faculty of Computer Science and Information Technology, Gunadarma University, Indonesia. He got his Ph.D dual degree in Combinatorics from Universite de BourgogneFrance under supervising of Prof. Vincent Vajnovszki and from Gunadarma University under supervising of Prof. Belawati Widjaja. Some of his papers were presented in some conference such as Words2005, CANT-2006, GASCom-2006, and some others are published in some journals or research reports such as CDMTCS-242 (2004), CDMTCS-276 (2006), The Computer Journal 60(5)-2007, Taru-DMSC 11(2)-2008.
A.B. Mutiara is a Professor of Computer Science. He is also Dean of Faculty of Computer Science and Information Technology, Gunadarma University, Indonesia.

The Hamming distance between the first and the last element of S2(123,132) is also 2, but the two terms are parted by (n-2) other terms since the first element is the image of 0n-1, namely (n-1)(n-2)...1n, while the last is the image of 10n-2, namely n(n-2)(n-3)...1(n-3). V. CONCLUDING REMARKS

Isomorphism between graph of Bn-1 and graph of Sn(123,132) is more simple than isomorphism between graph of Fn-1 and graph of Sn(123,132,213), where Fn-1 is the set of binary strings of length (n-1) having no 2 consecutive 1s. The constructive bijection between Fn-1 and Sn(123,132,213) showed by Simion-Schmidt [9]. There is no Hamiltonian cycle in this case, while Hamming distance between two consecutive elements of Sn(123,132,213), a Gray code for Sn(123,132,213), is also 2, as is showed by Juarna-Vajnovszki [3, 2].

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